Question

http://prntscr.com/9wietz

Answer 1

@TrojanPoem

Answer 2

@welshfella

Answer 3

refine first\[\frac{1}{u^3}=\frac{1}{125}\]multiply lowest common denominator \[\frac{1}{u^3}125u^3=\frac{1}{125}125u^3\] refine that \[u^3=125\] subtract 125 from both sides\[u^3-\left(125\right)=125-\left(125\right)\]factor\[u^3-125:\quad \left(u-5\right)\left(u^2+5u+25\right)\] solve for u \[\left(u-5\right)\left(u^2+5u+25\right)=0\]\[u-5=0\] solve\[u^2+5u+25\]by using the quadratic fromula \[u=\frac{-5+\sqrt{5^2-4\cdot \:25\cdot \:1}}{2\cdot \:1}=\frac{1}{2}\left(-5+5i\sqrt{3}\right)=\frac{5}{2}i\left(\sqrt{3}+i\right)\]\[u=\frac{-5-\sqrt{5^2-4\cdot \:25\cdot \:1}}{2\cdot \:1}=\frac{1}{2}\left(-5-5i\sqrt{3}\right)=-\frac{5}{2}i\left(\sqrt{3}-i\right)\]final soulutions are\[u=-\frac{5}{2}i\left(\sqrt{3}-i\right),\:u=\frac{5}{2}i\left(\sqrt{3}+i\right),\:u=5\]

Answer 4

sorry it took so long :/